# Dirac's delta in 3 dimensions: proof of $\nabla^2(\|\boldsymbol{x}-\boldsymbol{x}_0\|^{-1})=-4\pi\delta(\boldsymbol{x}-\boldsymbol{x}_0)$

If $T_f$ is a distribution, i.e. a linear functional, continuous according to the convergence defined here, defined on the space $K$ of the functions of class $C^\infty$ that are null outside a bounded interval (which is not the same for all functions), its derivative is defined as $$\frac{dT_f}{dx}(\varphi):=-T_f(\varphi')$$where $\varphi'$ is the derivative of $\varphi$. The symbolic writing $T_f(\varphi)=\int_{-\infty}^\infty f(x)\varphi(x)dx$ is often used to write such a functional, since, if $g$ is (Riemann or Lebesgue) integrable on every bounded interval, then $\int_{-\infty}^\infty g(x)\varphi(x)dx$ indeed is such a continuous functional. In this context, we can symbolically define the "derivative" $f'$, for any $T_f$, even if the symbolic writing $f$ does not refer to an integrable function, according to the expression$$\int_{-\infty}^\infty f'(x)\varphi(x)dx:=-\int_{-\infty}^\infty f(x)\varphi'(x)dx=:\frac{dT_f}{dx}(\varphi).$$

Let us come to my question. While studying physics, in particular the theory of electromagnetism and the derivation of the Biot-Savart law from Ampère's law, I always find the equality$$\nabla^2\left(\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}_0\|}\right)=-4\pi\delta(\boldsymbol{x}-\boldsymbol{x}_0)$$where $\nabla^2$ is the Laplacian$^1$. I suppose that, in the tridimensional case, with $\phi:\mathbb{R}^3\to\mathbb{R}$, $\int_{\mathbb{R}^3}\frac{\partial f(\boldsymbol{x})}{\partial x_i}\phi(\boldsymbol{x}) dx_1dx_2dx_3$ is analogously defined as $\frac{\partial T_f}{\partial x_i}(\varphi)$, which I suppose to be analogously defined, in turn, as $-\int_{\mathbb{R}^3}f(\boldsymbol{x})\frac{\partial \phi(\boldsymbol{x})}{\partial x_i} dx_1dx_2dx_3$, although I say I suppose because I have not found a rigourous definition of such derivatives on line nor in cartaceous texts; as to mathematical resources, I have studied Kolmogorov-Fomin's Элементы теории функций и функционального анализа, which only focuses on the monodimensional $\varphi:\mathbb{R}\to\mathbb{R}$ case. Once fixed a proper definition of such derivatives, how can it be proved that $\nabla^2(\|\boldsymbol{x}-\boldsymbol{x}_0\|^{-1})=-4\pi\delta(\boldsymbol{x}-\boldsymbol{x}_0)$?

$^1$ The link contains a derivation of $\nabla^2(\|\boldsymbol{x}-\boldsymbol{x}_0\|^{-1})=-4\pi\delta(\boldsymbol{x}-\boldsymbol{x}_0)$ (equations (18)-(24)), but I do not understand it: I would understand it if we could apply Gauss's divergence theorem at (20), but I know it for functions of class $C^1(\mathring{A})$, $\overline{V}\subset\mathring{A}$, only, while $\nabla\left(\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}_0\|}\right)$ is not even defined for $\boldsymbol{x}=\boldsymbol{x}_0$; the other derivation of the identity $\nabla^2(\|\boldsymbol{x}-\boldsymbol{x}_0\|^{-1})$ $=-4\pi\delta(\boldsymbol{x}-\boldsymbol{x}_0)$ that I have found uses a "weak limit", but it does not use the formal definiton of derivative of a distribution that I have written above. These are the two only references addressing what I am asking that I have managed to find.

• There's a derivation in the PDE book by Evans, section 2.2.1b (in the 1st edition at least), though he doesn't couch it in the language of distributions except in the remark at the end. You could also look at Section 4.5 of these notes. – epimorphic Jan 24 '16 at 16:30
• @epimorphic Thank you for your suggestions! – Self-teaching worker Jan 26 '16 at 10:21

You are right, the partial derivatives of distributions on higher-dimensional spaces are defined as you surmised,

$$\frac{\partial T}{\partial x_i} \colon \varphi \mapsto -T\biggl[\frac{\partial \varphi}{\partial x_i}\biggr],$$

more generally

$$D^{\alpha} T \colon \varphi \mapsto (-1)^{\lvert\alpha\rvert} T[D^{\alpha}\varphi]$$

for higher derivatives.

From that you obtain $(\nabla^2 T)[\varphi] = T[\nabla^2\varphi]$, and for the locally integrable function $f(x) = \lVert x-x_0\rVert^{-1}$ we therefore have

$$(\nabla^2 T_f)[\varphi] = T[\nabla^2\varphi] = \int_{\mathbb{R}^3} f(x)\cdot \nabla^2\varphi(x)\,dx_1\,dx_2\,dx_3.\tag{1}$$

Now choose $R > 0$ so large that $\operatorname{supp} \varphi \subset B_R(x_0)$, and choose $0 < \varepsilon < R$. Then

$$\int_{\mathbb{R}^3} f(x)\cdot \nabla^2\varphi(x)\,dx_1\,dx_2\,dx_3 = \lim_{\varepsilon \searrow 0} \int_{\varepsilon < \lVert x-x_0\rVert < R} f(x)\cdot \nabla^2\varphi(x)\,dx_1\,dx_2\,dx_3$$

since $f$ is locally integrable and $\nabla^2\varphi$ is continuous.

Away from $x_0$, $f$ is smooth, and a slightly tedious computation shows that $\nabla^2 f \equiv 0$ on $\mathbb{R}^3\setminus \{x_0\}$, so

$$\int_{\mathbb{R}^3} f(x)\cdot \nabla^2\varphi(x)\,dx_1\,dx_2\,dx_3 = \lim_{\varepsilon \searrow 0} \int_{\varepsilon < \lVert x-x_0\rVert < R} f(x)\cdot \nabla^2\varphi(x) - \varphi(x)\cdot \nabla^2 f(x)\,dx_1\,dx_2\,dx_3,$$

and to that integral you can apply Green's formula to obtain

\begin{align} \int_{\mathbb{R}^3} f(x)\cdot \nabla^2\varphi(x)\,dx_1\,dx_2\,dx_3 &= \lim_{\varepsilon \searrow 0}\; \Biggl(\int_{\lVert x-x_0\rVert = R} f(x)\frac{\partial \varphi}{\partial \nu}(x) - \varphi(x)\frac{\partial f}{\partial\nu}(x)\,dS\\ &\qquad\qquad + \int_{\lVert x-x_0\rVert = \varepsilon} f(x)\frac{\partial \varphi}{\partial \nu}(x) - \varphi(x)\frac{\partial f}{\partial\nu}(x)\,dS\Biggr)\\ &= \lim_{\varepsilon \searrow 0} \int_{\lVert x-x_0\rVert = \varepsilon} f(x)\frac{\partial \varphi}{\partial \nu}(x) - \varphi(x)\frac{\partial f}{\partial\nu}(x)\,dS, \end{align}

where $dS$ denotes the surface measure of the sphere, $\frac{\partial}{\partial\nu}$ the directional derivative in direction of the outer normal, and the first integral on the right hand side vanishes since $\varphi$ vanishes in a neighbourhood of the outer sphere. The outer normal on the sphere $\lVert x-x_0\rVert = \varepsilon$ is $-\frac{x-x_0}{\lVert x-x_0\rVert}$,

so we are left with

$$\frac{1}{\varepsilon}\int_{\lVert x-x_0\rVert = \varepsilon} f(x)\langle \nabla\varphi(x),x-x_0\rangle - \varphi(x)\langle \nabla f, x-x_0\rangle\,dS.$$

An easy estimate shows

$$\frac{1}{\varepsilon}\int_{\lVert x-x_0\rVert = \varepsilon} f(x)\langle\nabla\varphi(x),x-x_0\rangle\,dS \leqslant \frac{1}{\varepsilon^2}\int_{\lVert x-x_0\rVert = \varepsilon} \lVert\nabla\varphi\rVert\cdot\varepsilon\,dS \leqslant C\frac{4\pi\varepsilon^2}{\varepsilon} \xrightarrow{\varepsilon \searrow 0} 0.$$

For the other part of the integral, computing $\frac{\partial f}{\partial\nu}$ gives

$$\int_{\lVert x-x_0\rVert = \varepsilon} \varphi(x)\cdot \frac{1}{\varepsilon^2}\,dS,$$

and since $\varphi$ is continuous at $x_0$ we have

$$\lim_{\varepsilon\searrow 0} \int_{\lVert x-x_0\rVert = \varepsilon} \varphi(x)\cdot \frac{1}{\varepsilon^2}\,dS = 4\pi\varphi(x_0).$$

Collecting everything without mucking up the signs, we get

$$(\nabla^2 f)[\varphi] = \int_{\mathbb{R}^3} f(x)\cdot \nabla^2\varphi(x)\,dx_1\,dx_2\,dx_3 = -4\pi \varphi(x_0) = -4\pi\delta_{x_0}[\varphi].$$

That holds for all test functions $\varphi$, hence $\nabla^2 f = -4\pi\delta_{x_0}$.

• Thank you very much for your detailed answer! As to $\int_{\mathbb{R}^3} f(x)\cdot \nabla^2\varphi(x)\,dx_1\,dx_2\,dx_3$, I think it's intended to be a Lebesgue integral: am I right? There is only one thing that I don't understand: why precisely does the continuity of $\varphi$ entails that $\lim_{\varepsilon\searrow 0} \int_{\lVert x-x_0\rVert = \varepsilon} \frac{\varphi(x)}{\varepsilon^2}\,dS = 4\pi\varphi(x_0)$? I $\infty$-ly thank you. – Self-teaching worker Jan 25 '16 at 19:54
• The area of the sphere of radius $\varepsilon$ is $4\pi \varepsilon^2$, so that integral is $4\pi$ times the average of $\varphi$ on that sphere, and continuity of $\varphi$ (at $x_0$) makes that average converge to $\varphi(x_0)$. In formulae: $$\biggl\lvert \frac{1}{\varepsilon^2}\int_{S_{\varepsilon}} \varphi(x)\,dS - 4\pi \varphi(x_0)\biggr\rvert \leqslant \frac{1}{\varepsilon^2} \int_{S_{\varepsilon}} \lvert \varphi(x) - \varphi(x_0)\rvert\,dS \leqslant 4\pi \sup \{ \lvert \varphi(x) - \varphi(x_0)\rvert : \lVert x-x_0\rVert \leqslant \varepsilon\}.$$ – Daniel Fischer Jan 25 '16 at 20:14
• Continuity on the compact $S_\varepsilon$ implies uniform continuity... I heartily thank you! – Self-teaching worker Jan 25 '16 at 20:24

For a fixed $x_0$, define $G_{x_0}=|x-x_0|^{-1}$. Then $\nabla^2 G_{x_0}=0$ for $x \ne x_0$. If $\varphi$ is a compactly supported $C^{\infty}$ function on $\mathbb{R}^3$, then $$\nabla\cdot(G_{x_0}\nabla\varphi-\varphi \nabla G_{x_0})=G_{x_0}\nabla^2\varphi-\varphi\nabla^2G_{x_0}=G_{x_0}\nabla^2\varphi,\;\;\; x \ne x_0.$$ Integrate and apply the divergence theorem, keeping in mind that $\varphi$ vanishes outside a large sphere: \begin{align} \int_{\mathbb{R}^3} G_{x_0}\nabla^2\varphi dV &=\lim_{\epsilon\downarrow 0}\int_{|x-x_0|\ge \epsilon}G_{x_0}\nabla^2\varphi dV\\ &=-\lim_{\epsilon\downarrow 0}\int_{|x-x_0| \ge \epsilon}\nabla\cdot(\varphi \nabla G_{x_0} - G_{x_0}\nabla\varphi)dV \\ & =-\lim_{\epsilon\downarrow 0}\int_{|x-x_0|=\epsilon}\left(\varphi\frac{\partial G_{x_0}}{\partial n}-G_{x_0}\frac{\partial\varphi}{\partial n}\right)dS \\ & =-\lim_{\epsilon\downarrow 0}\int_{|x-x_0|=\epsilon}\varphi\frac{\partial G_{x_0}}{\partial n}dS \end{align} The normal derivative is in the inward direction on $|x-x_0|=\epsilon$ because the outward normal on $\epsilon \le |x-x_0| \le R$ is where this started. $G_{x_0}(x)=1/r$ for $|x-x_0|=r$. So you end up with a normal derivative of $G_{x_0}$ equal to $1/\epsilon^2$ on $|x-x_0|=\epsilon$. Therefore, you pick up a factor equal to the area of the unit sphere: $$\int_{\mathbb{R}^3} G_{x_0}\nabla^2\varphi dV = -\lim_{\epsilon\downarrow 0}\frac{1}{\epsilon^2}\int_{|x-x_0|=\epsilon}\varphi(x)dS = -4\pi\varphi(x_0).$$

• I heartily thank you for your very interesting answer! I have chosen the other answer because it's chronologically older and a bit more detailed, but I also really appreciate yours. – Self-teaching worker Jan 25 '16 at 20:25
• @Self-teachingDavide : I started the solution, had to leave the solution in progress; I came back and finished, only to find there was another solution already posted after I posted. I almost deleted, but decided to leave it. – Disintegrating By Parts Jan 25 '16 at 22:16